\(n_M=\frac{1,2}{M_M}\left(mol\right)\)
\(n_{H_2}=\frac{0,672}{22,4}=0,03\left(mol\right)\)
PTHH: \(2M+2nH_2O\rightarrow2M\left(OH\right)_n+nH_2\)
______\(\frac{0,06}{n}\)<----------------------------------0,03______(mol)
=> \(\frac{1,2}{M_M}=\frac{0,06}{n}\)
=> \(M_M=20n\) (g/mol)
Xét n = 1 => MM = 20 (g/mol) => loại
Xét n = 2 => MM = 40 (g/mol) => M là Ca